Question: Divide the following complex numbers. $ \dfrac{-20+4i}{3+2i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3-2i}$ $ \dfrac{-20+4i}{3+2i} = \dfrac{-20+4i}{3+2i} \cdot \dfrac{{3-2i}}{{3-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-20+4i) \cdot (3-2i)} {(3+2i) \cdot (3-2i)} = \dfrac{(-20+4i) \cdot (3-2i)} {3^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-20+4i) \cdot (3-2i)} {(3)^2 - (2i)^2} = $ $ \dfrac{(-20+4i) \cdot (3-2i)} {9 + 4} = $ $ \dfrac{(-20+4i) \cdot (3-2i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-20+4i}) \cdot ({3-2i})} {13} = $ $ \dfrac{{-20} \cdot {3} + {4} \cdot {3 i} + {-20} \cdot {-2 i} + {4} \cdot {-2 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{-60 + 12i + 40i - 8 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{-60 + 12i + 40i + 8} {13} = \dfrac{-52 + 52i} {13} = -4+4i $